partial differentiation question

[denki23]

pg. 233-4 in Horowitz and Hill's The Art of Electronics discusses feedback with finite-gain amplifiers. A formula for closed loop voltage gain is derived. G = A / (1+ AB). A is the open loop gain of the amplifier and varies with frequency. The book then says "it is easy to show, by taking the partial derivative of G with respect to A (δG/δA). that relative variations in the open-loop gain are reduced by the desensitivity: ∆G/G = (1/1+AB)*(∆A/A)." (desensitivity is 1+AB)

I don't understand how the book arrived at that last expression. I tried taking partial derivative with respect to A and end up with a much messier expression after use of product and quotient rules. Appreciate any help.

 

[DrPhil]

pg. 233-4 in Horowitz and Hill's The Art of Electronics discusses feedback with finite-gain amplifiers. A formula for closed loop voltage gain is derived. G = A / (1+ AB). A is the open loop gain of the amplifier and varies with frequency. The book then says "it is easy to show, by taking the partial derivative of G with respect to A (δG/δA). that relative variations in the open-loop gain are reduced by the desensitivity: ∆G/G = (1/1+AB)*(∆A/A)." (desensitivity is 1+AB)

I don't understand how the book arrived at that last expression. I tried taking partial derivative with respect to A and end up with a much messier expression after use of product and quotient rules. Appreciate any help.

Partial derivative with respect to A just means B is a constant.

\(\displaystyle \displaystyle \Delta G = \dfrac{\partial G}{\partial A} \Delta A = \left[ \dfrac{1}{1 + AB} - \dfrac{AB}{(1 + AB)^2} \right] \Delta A = \dfrac{\Delta A}{(1 + AB)^2} \)

\(\displaystyle \displaystyle \dfrac{\Delta G}{G} = \left[\dfrac{\Delta A}{(1 + AB)^2}\right]\left[\dfrac{1+AB}{A}\right] = \dfrac{1}{1+AB} \times \dfrac{\Delta A}{A} \)

 

[denki23]

Now it makes sense. Thank you.